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Published byDimitri Erwin Modified about 1 year ago

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Shortest paths and transitive closure Data structure 2002/12/4

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Is there a path? How short it can be? Single source/ All destinations –Nonnegative edge costs –General weights All-pairs shortest path Transitive closure

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Single source all destinations Dijkstra’s algorithm A spanning tree again For nonnegative edge costs (Why?) Start from a vertex v, greedy method dist[w]: the shortest length to w through S v u w length( ) dist[w] dist[u]

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shortest() Void shortestpath(int v, int cost[][MAX_VERTICES], int dist [], int n, int found[]) { int i,u,w; for (i=0;i

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An example San Francisco Los Angeles Denver Chicago Boston New York Miami New Orleans

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Single source all destinations BellmanFord algorithm For general weights Path has at most n-1 edges, otherwise… dist k [u]: from v to u, has at most k edges

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BellmanFord() Void BellmanFord(int n, int v) { int i,k; for (i =0;i

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All-Pairs shortest paths 執行 n 次 single source all destinations algo. Dynamic programming strategy – 利用 recursive formula 來表示 – 好好地 implement recursive formula, 用 table 輔助 A k [i][j] ≡ shortest length from i to j through no intermediate vertex greater than k A -1 [i][j] = length[i][j] A k [i][j] = min{A k-1 [i][j], A k-1 [i][k]+ A k-1 [k][j], k≥0

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AllLengths() Void AllLength(int n) { int i,j,k; for(i=0;i

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Transitive closure Definition: transitive closure matrix, A + –A + [i][j] = 1, if there’s a path of length > 0 from i to j –A + [i][j] = 0, otherwise Definition: reflexive transitive closure matrix, A* –A*[i][j] = 1, if there’s a path of length >= 0 from i to j –A*[i][j] = 0, otherwise O(n^3), by AllLengths() O(n^2), an undirected graph, by connected check

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